The integration for the area of the extremal surface can be performed by using \(\zeta \)  as an independent variable. The area becomes [1]}, [2]}
\({\rm Area}(\gamma _A)\equiv I(\epsilon )=2{\tilde{L}}^2\frac{1}{\zeta _0^2}\int _{\epsilon _\zeta (T)/\zeta _0}^1dy\frac{1}{y^3\sqrt{1-y^6}},\) 
